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MathGroup Archive 2005

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Re: Is this possible? Residue computation leads to complexInfinity?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg60725] Re: [mg60719] Is this possible? Residue computation leads to complexInfinity?
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Mon, 26 Sep 2005 01:36:04 -0400 (EDT)
  • Reply-to: hanlonr at cox.net
  • Sender: owner-wri-mathgroup at wolfram.com

Your Residue expression is unevaluated (Residue appears in output) since you 
left out part of the Residue syntax. However, your argument was evaluated to 
ComplexInfinity so z must already have had a value.

Clear[z];
FullSimplify[
  Residue[(I*z)/(1-(2+4*a)*z^2+z^4),
    {z,Sqrt[a+1]-Sqrt[a]}],a>0]

(I*(Sqrt[1 + 1/a] - 1))/(8*(-a + Sqrt[a^2 + a] - 1))


Bob Hanlon

> 
> From: "kiki" <lunaliu3 at yahoo.com>
To: mathgroup at smc.vnet.net
> Date: 2005/09/25 Sun AM 02:36:33 EDT
> Subject: [mg60725] [mg60719]  Is this possible? Residue computation leads to 
complexInfinity?
> 
> In[15]:=
> \!\(FullSimplify[Residue[\(\[ImaginaryI]\ z\)\/\(1 - \((2 + 4\ a)\)\ z\^2 + 
> z\
> \^4\), \@\(a + 1\) - \@a], \ a > 0]\)
> 
> Out[15]=
> \!\(Residue[ComplexInfinity, \(-\@a\) + \@\(1 + a\)]\) 
> 
> 
> 


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